MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5b Molecular Geometry Hybridization
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5b Molecular Geometry HybridizationQuestion 1 of 20

In a screen for volatile anesthetic analogs, a lab compares two isomers with formula C2H3Cl. Compound X shows an IR band near 3300 cm−1^{-1} and forms a precipitate with AgNO3_3 in ethanol only after prolonged heating; Compound Y reacts rapidly. Compound X is assigned as chloroethyne (HC≡C–Cl). Which hybridization state is most consistent with the bonding at the carbon directly bonded to Cl in HC≡C–Cl?

sp3^3, because the carbon forms four σ\sigma bonds
sp2^2, because the carbon participates in one π\pi bond
sp, because the carbon is part of a C≡C triple bond
Unhybridized p, because the carbon uses only p orbitals for bonding
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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5b Molecular Geometry Hybridization

Practice 5b Molecular Geometry Hybridization in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 5b Molecular Geometry Hybridization, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

In a screen for volatile anesthetic analogs, a lab compares two isomers with formula C2H3Cl. Compound X shows an IR band near 3300 cm−1^{-1} and forms a precipitate with AgNO3_3 in ethanol only after prolonged heating; Compound Y reacts rapidly. Compound X is assigned as chloroethyne (HC≡C–Cl). Which hybridization state is most consistent with the bonding at the carbon directly bonded to Cl in HC≡C–Cl?

  1. sp3^3, because the carbon forms four σ\sigma bonds
  2. sp2^2, because the carbon participates in one π\pi bond
  3. sp, because the carbon is part of a C≡C triple bond (correct answer)
  4. Unhybridized p, because the carbon uses only p orbitals for bonding
Explanation: This question tests the determination of hybridization based on bonding in alkynes. Hybridization is determined by the number of electron domains around an atom, where two domains indicate sp hybridization for linear geometry. In chloroethyne (HC≡C–Cl), the carbon bonded to Cl has two electron domains: one from the C≡C triple bond (counting as one domain) and one from the C–Cl single bond. The sp hybridization is consistent because it allows for two sigma bonds and two pi bonds in the triple bond, resulting in linear geometry at that carbon. Choice A fails due to the misconception that a triple bond counts as three separate domains, but it is one domain for hybridization purposes. To verify hybridization, count the number of sigma bonds and lone pairs as electron domains. Additionally, confirm by noting that sp hybridized carbons in terminal alkynes exhibit linear R–C≡C arrangements.

Question 2

A medicinal chemistry group compares two amine-containing buffers for maintaining pH during enzyme assays. One candidate is trimethylamine, N(CH3_3)3_3, which shows a strong dipole moment consistent with a nonplanar nitrogen center. Based on VSEPR theory, which molecular geometry is expected around the nitrogen atom in neutral trimethylamine?

  1. Trigonal planar, with sp2^2 hybridization and 120° bond angles
  2. Tetrahedral, because four atoms are bonded to nitrogen
  3. Trigonal pyramidal, because three bonding pairs and one lone pair occupy four electron domains (correct answer)
  4. Linear, because nitrogen forms three equivalent π\pi bonds to carbon
Explanation: This question tests VSEPR theory application to determine molecular geometry. Trimethylamine N(CH₃)₃ has a nitrogen atom with three bonding pairs (to the three methyl groups) and one lone pair, giving four total electron domains. According to VSEPR theory, four electron domains arrange tetrahedrally, but with one position occupied by a lone pair, the molecular geometry becomes trigonal pyramidal. The presence of the lone pair creates an asymmetric charge distribution, resulting in the observed dipole moment and nonplanar nitrogen center. Choice B incorrectly confuses electron-domain geometry with molecular geometry - while there are four electron domains arranged tetrahedrally, only three are bonds, making the molecular shape pyramidal. When applying VSEPR, always distinguish between electron-domain geometry (including lone pairs) and molecular geometry (considering only atoms).

Question 3

A biochemist compares carbon centers in lactate (CH3_3–CH(OH)–COO−^-) during enzyme binding. The middle carbon (CH) is bonded to CH3_3, OH, COO−^-, and H, and is observed to be a stereocenter. Which geometry and hybridization are most consistent for this middle carbon?

  1. Trigonal planar, sp2^2, because it is adjacent to a carbonyl-containing group
  2. Tetrahedral, sp3^3, because it has four σ\sigma bonds (four electron domains) (correct answer)
  3. Linear, sp, because stereocenters require 180° bond angles
  4. Bent, sp3^3, because one lone pair on carbon distorts the geometry
Explanation: This question tests geometry and hybridization of tetrahedral carbon centers. The middle carbon in lactate is bonded to four different groups (CH₃, OH, COO⁻, and H), creating four electron domains that require sp³ hybridization. Four sp³ hybrid orbitals arrange tetrahedrally with bond angles of approximately 109.5°, and this tetrahedral geometry allows the carbon to be a stereocenter (chiral) when bonded to four different groups. Choice A incorrectly suggests sp² hybridization based on proximity to a carbonyl, but each carbon's hybridization depends on its own electron domains, not neighboring groups. Choice C misunderstands that stereocenters require tetrahedral, not linear, geometry. For any carbon with four single bonds, the geometry is always tetrahedral with sp³ hybridization, regardless of what groups are attached.

Question 4

During oxidative DNA damage, guanine can be converted to 8-oxoguanine, altering hydrogen-bonding patterns. In both guanine and 8-oxoguanine, the carbonyl carbon participates in resonance with adjacent heteroatoms and is observed to be approximately planar by NMR coupling analysis. Which hybridization state is most consistent for a carbonyl carbon in these bases?

  1. sp3^3, because carbon is bonded to oxygen and two other atoms (three total)
  2. sp2^2, because the carbon has three electron domains and one unhybridized p orbital for the π\pi bond (correct answer)
  3. sp, because the C=O bond requires two unhybridized p orbitals
  4. sp3^3d, because carbonyls require expanded octets to maintain planarity
Explanation: This question tests hybridization of carbonyl carbons in biological molecules. A carbonyl carbon (C=O) has three electron domains: one double bond to oxygen (counted as one domain) and two single bonds to other atoms. Three electron domains require sp² hybridization, which creates three hybrid orbitals in a trigonal planar arrangement and leaves one unhybridized p orbital for the π bond component of the C=O double bond. The observed planarity by NMR coupling analysis confirms sp² hybridization, as this geometry allows all atoms bonded to the carbonyl carbon to lie in the same plane. Choice A incorrectly counts bonds instead of electron domains, while choice C misunderstands that only one p orbital is needed for the π bond in a double bond. For carbonyl carbons, the consistent pattern is sp² hybridization with trigonal planar geometry.

Question 5

A lab measures volatility of two small molecules used as anesthetic adjuncts: ethanol (CH3_3CH2_2OH) and dimethyl ether (CH3_3OCH3_3). Both contain an oxygen atom with two lone pairs. Based on VSEPR theory, what is the expected molecular geometry around the oxygen atom in each compound?

  1. Linear, because oxygen has two bonds
  2. Trigonal planar, because oxygen has three electron domains (two bonds and one lone pair)
  3. Bent, because oxygen has four electron domains (two bonds and two lone pairs) (correct answer)
  4. Tetrahedral, because molecular geometry equals electron-domain geometry when lone pairs are present
Explanation: This question tests VSEPR theory for determining molecular geometry around oxygen atoms. In both ethanol and dimethyl ether, the oxygen atom has two bonding pairs (C-O and O-H in ethanol; two C-O bonds in ether) and two lone pairs, giving four total electron domains. Four electron domains arrange tetrahedrally, but with two positions occupied by lone pairs, the molecular geometry is bent (angular). The bond angle is approximately 104.5°, slightly compressed from the tetrahedral angle due to lone pair repulsion. Choice A incorrectly counts only three electron domains, while choice D confuses electron-domain geometry with molecular geometry. Remember that for VSEPR, molecular geometry describes only the arrangement of atoms, not lone pairs, so four domains with two lone pairs always gives a bent shape.

Question 6

A formulation scientist evaluates urea, CO(NH2_2)2_2, as a protein denaturant. Spectroscopic data indicate the carbonyl carbon is planar with its three attached atoms (O, N, N), consistent with resonance between C=O and C–N bonds. Based on this, which is the most consistent hybridization of the carbonyl carbon in urea?

  1. sp2^2, because three electron domains at carbon give trigonal planar electron geometry (correct answer)
  2. sp3^3, because carbon is single-bonded to two nitrogens and must be tetrahedral
  3. sp, because planarity requires linear hybrid orbitals
  4. sp3^3d, because resonance requires five equivalent electron domains
Explanation: This question tests hybridization in molecules with resonance structures. In urea CO(NH₂)₂, the carbonyl carbon is bonded to one oxygen and two nitrogen atoms, giving three electron domains. Three electron domains require sp² hybridization, resulting in a trigonal planar arrangement that allows all four atoms (C, O, N, N) to lie in the same plane. The sp² hybridization provides an unhybridized p orbital that participates in resonance between the C=O double bond and partial double bond character in the C-N bonds. Choice B incorrectly assumes sp³ hybridization based on single bonds, ignoring the resonance that creates partial double bond character. When analyzing molecules with resonance, count electron domains based on the resonance hybrid structure, not individual resonance forms.

Question 7

A biochemistry lab prepares bicarbonate buffer by dissolving NaHCO3_3 in water. To interpret the dominant intermolecular forces of the bicarbonate anion, the lab models the carbon center as bonded to three oxygens (one double bond and two single bonds, with resonance). Based on VSEPR theory for the carbon atom in HCO3−_3^-, which geometry is expected around carbon?

  1. Tetrahedral, because resonance implies four electron domains around carbon
  2. Trigonal planar, because carbon has three electron domains and no lone pairs (correct answer)
  3. Trigonal pyramidal, because carbon has one lone pair in the anion
  4. Linear, because a π\pi bond forces 180° bond angles
Explanation: This question tests VSEPR geometry prediction for the bicarbonate ion's carbon center. In HCO₃⁻, the carbon atom is bonded to three oxygen atoms with resonance structures showing one C=O double bond and two C-O single bonds (though all three bonds are equivalent due to resonance). Regardless of resonance, the carbon has three electron domains (three sigma bonds) and no lone pairs, resulting in a trigonal planar geometry according to VSEPR theory. This corresponds to sp² hybridization at carbon, with bond angles of 120°. Choice A incorrectly interprets resonance as creating four electron domains. Choice C wrongly assigns a lone pair to carbon, which would violate its valence. Choice D suggests linear geometry, impossible with three substituents. The transferable principle is that resonance affects bond order and electron distribution but doesn't change the number of electron domains for geometry determination.

Question 8

In a headspace GC-MS method for monitoring acetone in exhaled breath (a diabetes-adjacent ketone marker), a technician compares acetone (CH3_3–C(=O)–CH3_3) to isopropanol (CH3_3–CH(OH)–CH3_3) to rationalize differences in intermolecular interactions. Based on the bonding at the carbonyl carbon in acetone, which hybridization state is most consistent with the observed planar carbonyl group and the presence of a π\pi bond?

  1. sp3^3, because four electron domains surround the carbonyl carbon
  2. sp, because two electron domains create a linear carbonyl carbon
  3. sp2^2, because three electron domains form a trigonal planar arrangement with one unhybridized p orbital for the π\pi bond (correct answer)
  4. dsp2^2, because a double bond requires d-orbital participation in second-row elements
Explanation: This question tests understanding of hybridization states and molecular geometry around a carbonyl carbon. In carbonyl compounds, the carbon atom forms three sigma bonds (one C=O double bond consists of one sigma and one pi bond, plus two C-C single bonds) and has no lone pairs, giving it three electron domains. According to VSEPR theory, three electron domains arrange in a trigonal planar geometry with 120° bond angles, which requires sp² hybridization. The sp² hybridization uses three hybrid orbitals for sigma bonding, leaving one unhybridized p orbital perpendicular to the plane for pi bond formation with oxygen. Choice A incorrectly counts four electron domains by double-counting the pi bond, while choice B suggests only two domains, ignoring the methyl groups. Choice D incorrectly invokes d-orbital participation, which is not possible for carbon. To verify hybridization, count sigma bonds plus lone pairs on the atom of interest - for carbonyl carbon, this gives 3 + 0 = 3, confirming sp² hybridization.

Question 9

A formulation chemist studies hydrogen bonding in aqueous ammonia versus ammonium chloride solutions at the same total nitrogen concentration. In NH3_3, nitrogen is bonded to three hydrogens and has one lone pair. Based on VSEPR theory, which combination best describes the electron-domain geometry and the molecular geometry of NH3_3 around nitrogen?

  1. Electron-domain: trigonal planar; molecular: trigonal planar
  2. Electron-domain: tetrahedral; molecular: trigonal pyramidal (correct answer)
  3. Electron-domain: tetrahedral; molecular: tetrahedral
  4. Electron-domain: trigonal bipyramidal; molecular: seesaw
Explanation: This question tests the distinction between electron-domain geometry and molecular geometry in VSEPR theory. For ammonia (NH₃), the nitrogen atom has four electron domains: three N-H bonds and one lone pair. Four electron domains arrange in a tetrahedral electron-domain geometry to minimize repulsion. However, the molecular geometry only considers the positions of atoms, not lone pairs, resulting in a trigonal pyramidal shape with bond angles slightly less than 109.5° due to lone pair repulsion. This corresponds to sp³ hybridization at nitrogen. Choice A incorrectly assigns trigonal planar electron-domain geometry, which would require only three electron domains. Choice C fails to distinguish between electron-domain and molecular geometry. Choice D suggests five electron domains, which is impossible for nitrogen. The key principle is that electron-domain geometry includes all electron domains (bonds + lone pairs), while molecular geometry only describes atom positions.

Question 10

In a protein crosslinking assay, a lysine side-chain amine attacks a carbonyl carbon to form a tetrahedral intermediate before re-forming a carbonyl during product formation. Considering the carbonyl carbon of an aldehyde/ketone before nucleophilic attack, which molecular geometry is most consistent with its orbital description and bonding?

  1. Trigonal planar, consistent with sp2^2 hybridization at the carbonyl carbon (correct answer)
  2. Tetrahedral, consistent with sp3^3 hybridization at the carbonyl carbon
  3. Trigonal pyramidal, consistent with sp3^3 hybridization and one lone pair on carbon
  4. Linear, consistent with sp hybridization at the carbonyl carbon
Explanation: This question tests understanding of carbonyl carbon geometry before nucleophilic attack. In aldehydes and ketones, the carbonyl carbon forms three electron domains: one double bond to oxygen (counting as one domain) and two single bonds to other groups. With three electron domains and no lone pairs on carbon, VSEPR theory predicts a trigonal planar geometry with bond angles of 120°. This geometry corresponds to sp² hybridization, where three sp² hybrid orbitals form sigma bonds in a plane, and the unhybridized p orbital forms the pi bond with oxygen. Choice B incorrectly suggests tetrahedral geometry, which would require four electron domains. Choice C impossibly assigns a lone pair to carbon in a neutral carbonyl. Choice D suggests linear geometry, which requires only two electron domains. The key insight is that the pi bond in C=O doesn't create a separate electron domain - the entire double bond counts as one domain for geometry determination.

Question 11

A medicinal chemistry group evaluates membrane permeability of a tertiary amine that can be protonated in physiological buffers. The nitrogen center is bonded to three carbon substituents and has one lone pair when unprotonated, but after protonation it is bonded to four substituents with no lone pair. Based on VSEPR theory, which molecular geometry is expected around nitrogen after protonation?

  1. Trigonal pyramidal, because a lone pair still occupies one vertex
  2. Tetrahedral, because four bonding domains surround nitrogen (correct answer)
  3. Trigonal planar, because protonation converts nitrogen to sp2^2
  4. Linear, because nitrogen forms four equivalent bonds by sp hybridization
Explanation: This question tests molecular geometry determination using VSEPR theory for a protonated amine. When a tertiary amine (NR₃) becomes protonated to form an ammonium ion (NR₃H⁺), the nitrogen atom transitions from having three bonds plus one lone pair to having four bonds with no lone pairs. According to VSEPR theory, four electron domains (all bonding) around a central atom arrange themselves in a tetrahedral geometry with bond angles of approximately 109.5°. The molecular geometry equals the electron-domain geometry when no lone pairs are present, so the protonated nitrogen exhibits tetrahedral geometry. Choice A incorrectly suggests a lone pair remains after protonation, while choice C wrongly claims sp² hybridization for a four-coordinate nitrogen. Choice D impossibly suggests linear geometry for four substituents. A key check is recognizing that protonation adds a fourth bond without changing the total electron count around nitrogen - it simply converts the lone pair into a bonding pair.

Question 12

A neurochemistry lab compares CO2_2 and SO2_2 as gaseous modulators in experimental chambers. Both are triatomic, but CO2_2 shows no permanent dipole moment while SO2_2 does. Assuming idealized structures, which geometry around the central atom is most consistent with CO2_2 having no net dipole?

  1. Bent, because two double bonds create unequal electron repulsions
  2. Linear, because two electron domains around carbon are arranged 180° apart (correct answer)
  3. Trigonal planar, because carbon uses sp2^2 hybridization in CO2_2
  4. Tetrahedral, because carbon has four total bonding pairs in two double bonds
Explanation: This question tests molecular geometry and its relationship to molecular polarity. Carbon dioxide (CO₂) has a central carbon atom bonded to two oxygen atoms through double bonds. The carbon has two electron domains (each double bond counts as one domain) and no lone pairs, resulting in a linear geometry with a 180° O-C-O bond angle according to VSEPR theory. This linear arrangement causes the two C=O bond dipoles to point in opposite directions and cancel out, resulting in no net dipole moment for the molecule. Choice A incorrectly suggests bent geometry, which would require lone pairs on carbon. Choice C suggests trigonal planar geometry, requiring three electron domains. Choice D misinterprets double bonds as creating four domains. The contrast with SO₂ (which is bent due to a lone pair on sulfur) reinforces that molecular symmetry determines overall polarity. A key check is that linear geometry with identical terminal atoms always produces a nonpolar molecule.

Question 13

A biochemistry lab monitors conversion of a primary alcohol to an aldehyde (R–CH2OH → R–CHO) during an oxidation step in a synthetic pathway. Focusing on the carbon atom that becomes the carbonyl carbon (the carbon bonded to O in both functional groups), which change in hybridization is most consistent with this reaction?

  1. sp3 → sp2 (correct answer)
  2. sp2 → sp3
  3. sp3 → sp
  4. sp → sp2
Explanation: This question tests understanding of hybridization changes during oxidation reactions. In a primary alcohol (R–CH2OH), the carbon bearing the OH group has four single bonds (two to H, one to R, one to O), giving it four electron domains and sp3 hybridization with tetrahedral geometry. When oxidized to an aldehyde (R–CHO), this same carbon now has three electron domains (one double bond to O, one bond to R, one to H), adopting sp2 hybridization with trigonal planar geometry. The change from sp3 to sp2 reflects the loss of one electron domain as two C-H bonds are replaced by one C=O double bond. A common error is reversing the hybridization change (choice B) or thinking the carbon becomes sp hybridized (choice C). To track hybridization changes, count electron domains before and after the reaction: alcohols have sp3 carbons while carbonyl carbons are sp2.

Question 14

In a headspace GC-MS method used to quantify ethanol produced by yeast fermentation, a technician compares ethanol (CH3CH2OH) to dimethyl ether (CH3OCH3), an isomeric compound with the same formula. Assuming the oxygen atom is the central atom in each case, which molecular geometry is most consistent with the oxygen's electron-domain arrangement in both molecules under standard conditions?

  1. Linear molecular geometry at O because there are two sigma bonds and no lone pairs
  2. Trigonal planar molecular geometry at O because there are three electron domains around O
  3. Bent molecular geometry at O because there are four electron domains around O (correct answer)
  4. Tetrahedral molecular geometry at O because O forms four sigma bonds
Explanation: This question tests understanding of molecular geometry around oxygen atoms using VSEPR theory. In both ethanol (CH3CH2OH) and dimethyl ether (CH3OCH3), the oxygen atom has two sigma bonds and two lone pairs, giving it four electron domains total. According to VSEPR theory, four electron domains arrange tetrahedrally in space to minimize electron-pair repulsion. However, when determining molecular geometry, we only consider the positions of atoms, not lone pairs, resulting in a bent molecular shape around oxygen. A common misconception is thinking that two bonds automatically mean linear geometry (choice A), but this ignores the lone pairs that also occupy space around the oxygen. To verify molecular geometry, always count all electron domains (bonds + lone pairs) first for electron-domain geometry, then consider only bonded atoms for molecular geometry.

Question 15

A formulation scientist studies ammonia (NH3) as a potential base to adjust pH in a buffered biological assay. Based on VSEPR theory for the nitrogen atom, which molecular geometry is expected for NH3, and what is the corresponding electron-domain geometry?

  1. Trigonal planar molecular geometry; trigonal planar electron-domain geometry
  2. Tetrahedral molecular geometry; tetrahedral electron-domain geometry
  3. Trigonal pyramidal molecular geometry; tetrahedral electron-domain geometry (correct answer)
  4. Bent molecular geometry; trigonal planar electron-domain geometry
Explanation: This question tests the distinction between electron-domain geometry and molecular geometry for ammonia. The nitrogen atom in NH3 has three N-H bonds and one lone pair, giving it four electron domains total. Four electron domains arrange tetrahedrally to minimize repulsion, so the electron-domain geometry is tetrahedral. However, molecular geometry only considers atom positions, not lone pairs, so with three atoms bonded to nitrogen in a tetrahedral arrangement with one position occupied by a lone pair, the molecular geometry is trigonal pyramidal. A common error is confusing electron-domain geometry with molecular geometry (choice A), forgetting that lone pairs affect shape but aren't included in molecular geometry descriptions. The lone pair on nitrogen also makes NH3 polar and enables its function as a base. Always distinguish between total electron domains (for electron-domain geometry) and bonded atoms only (for molecular geometry).

Question 16

A formulation chemist evaluates sulfur dioxide (SO2_2) as a sterilant gas. Spectroscopy indicates sulfur has two S–O bonding regions and one lone pair (with resonance). Based on VSEPR, which molecular shape is most consistent for SO2_2?

  1. Linear
  2. Trigonal planar
  3. Bent (correct answer)
  4. Tetrahedral
Explanation: This question tests VSEPR prediction of bent geometry in molecules with lone pairs. VSEPR arranges three electron domains trigonal planar, but one lone pair yields bent molecular geometry. In SO₂, sulfur has two bonding domains to oxygen and one lone pair, with resonance. Bent geometry is consistent because the lone pair occupies space, bending the O–S–O angle to about 120°. Choice A fails by assuming linear, ignoring the lone pair's effect on shape. Include lone pairs in electron domain counts for accurate molecular geometry. Verify by noting bond angles less than 120° due to lone pair repulsion.

Question 17

A lab evaluates boron trifluoride (BF3_3) as a Lewis acid catalyst for a protecting-group step. In BF3_3, boron is bonded to three fluorines and has no lone pairs on boron. Based on VSEPR, which geometry is expected around boron?

  1. Trigonal planar (correct answer)
  2. Trigonal pyramidal
  3. Tetrahedral
  4. Linear
Explanation: This question tests geometry in electron-deficient compounds like BF₃. VSEPR predicts trigonal planar for three electron domains with no lone pairs. In BF₃, boron has three bonds to fluorine and an empty p orbital, but no lone pairs. Trigonal planar geometry is consistent because it minimizes repulsion among the three bonding domains. Choice B fails by suggesting pyramidal, which requires a lone pair not present on boron. Account for exceptions in electron-deficient atoms when applying VSEPR. Verify by noting 120° bond angles in spectroscopic data.

Question 18

A lab compares the polarity of CO2_2 and SO2_2 when dissolved in aqueous media. CO2_2 has two electron domains around carbon (two double bonds) and no lone pairs on carbon. Based on VSEPR, which geometry at carbon best accounts for CO2_2 being nonpolar despite polar C=O bonds?

  1. Bent, causing dipoles to add
  2. Linear, causing dipoles to cancel (correct answer)
  3. Trigonal planar, causing partial cancellation
  4. Tetrahedral, because carbon has four electron domains
Explanation: This question evaluates how geometry affects molecular polarity in linear molecules. VSEPR predicts linear geometry for two electron domains, allowing symmetric dipole cancellation. In CO₂, carbon has two double bonds, counting as two domains with no lone pairs. Linear geometry is consistent because it places polar C=O bonds in opposition, canceling the net dipole. Choice A fails by suggesting bent, which would make CO₂ polar like SO₂, but CO₂ is nonpolar. Assess polarity by considering geometry and vector sum of bond dipoles. Verify by comparing to bent molecules where dipoles add constructively.

Question 19

In an enzyme mechanism study, a carboxylate side chain (–COO−^-) is modeled with two equivalent C–O bonds due to resonance. The central carbon is attached to two oxygens and one substituent R. Which hybridization at the carboxylate carbon is most consistent with the observed bond equivalence and planarity?

  1. sp3^3
  2. sp2^2 (correct answer)
  3. sp
  4. sp3^3d
Explanation: This question evaluates hybridization in carboxylate ions with resonance. Hybridization for three electron domains is sp², supporting trigonal planar geometry. In –COO⁻, the carbon has three domains: two to oxygen (resonance-averaged) and one to R. Sp² hybridization is consistent because it enables pi delocalization, making C–O bonds equivalent and planar. Choice A fails by assuming sp³, which would imply non-equivalent bonds without resonance. Examine resonance for effective domain count and hybridization. Verify by noting bond lengths intermediate between single and double.

Question 20

To tune membrane permeability, a lab compares trimethylamine (N(CH3_3)3_3) with its protonated form, trimethylammonium (N(CH3_3)3_3H+^+). At physiological pH, the protonated form predominates. Based on VSEPR, which change in molecular geometry at nitrogen is expected upon protonation?

  1. Trigonal pyramidal to tetrahedral (correct answer)
  2. Tetrahedral to trigonal pyramidal
  3. Trigonal planar to trigonal pyramidal
  4. Linear to trigonal planar
Explanation: This question evaluates changes in molecular geometry upon protonation using VSEPR theory. VSEPR predicts geometry based on electron domains, where four domains with one lone pair give trigonal pyramidal, and four bonding pairs give tetrahedral. In trimethylamine (N(CH₃)₃), nitrogen has three bonds and one lone pair, becoming four bonds in the protonated form (N(CH₃)₃H⁺). The change from trigonal pyramidal to tetrahedral is consistent because protonation converts the lone pair to a bonding pair, altering the molecular shape. Choice B fails due to reversing the order, mistakenly assuming deprotonation instead of protonation. To check geometry changes, compare electron domain counts before and after reaction. Verify by noting that pyramidal shapes invert rapidly, while tetrahedral are more rigid.